# Integral of (37-11x)/((-1x-2+x^2)(x-3))

## \int\frac{37-11x}{\left(x^2-x-2\right)\left(x-3\right)}dx

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$\ln\left|x-3\right|+4\ln\left|1+x\right|-5\ln\left|x-2\right|+C_0$

## Step by step solution

Problem

$\int\frac{37-11x}{\left(x^2-x-2\right)\left(x-3\right)}dx$
1

Factor the trinomial $\left(-2-x+x^2\right)$ finding two numbers that multiply to form $-2$ and added form $-1$

$\begin{matrix}\left(-2\right)\left(1\right)=-2\\ \left(-2\right)+\left(1\right)=-1\end{matrix}$
2

Thus

$\int\frac{37-11x}{\left(x-3\right)\left(1+x\right)\left(x-2\right)}dx$
3

Using partial fraction decomposition, the fraction $\frac{37-11x}{\left(x-3\right)\left(1+x\right)\left(x-2\right)}$ can be rewritten as

$\frac{37-11x}{\left(x-3\right)\left(1+x\right)\left(x-2\right)}=\frac{A}{x-3}+\frac{B}{1+x}+\frac{C}{x-2}$
4

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x-3\right)\left(1+x\right)\left(x-2\right)$

$37-11x=\left(\frac{A}{x-3}+\frac{B}{1+x}+\frac{C}{x-2}\right)\left(x-3\right)\left(1+x\right)\left(x-2\right)$
5

Multiplying polynomials

$37-11x=\frac{A\left(x-3\right)\left(1+x\right)\left(x-2\right)}{x-3}+\frac{B\left(x-3\right)\left(1+x\right)\left(x-2\right)}{1+x}+\frac{C\left(x-3\right)\left(1+x\right)\left(x-2\right)}{x-2}$
6

Simplifying

$37-11x=A\left(1+x\right)\left(x-2\right)+B\left(x-3\right)\left(x-2\right)+C\left(x-3\right)\left(1+x\right)$
7

Expand the polynomial

$37-11x=-3C+Cx-2A-2Ax+Ax+Ax\cdot x+6B-2Bx-3Bx+Bx\cdot x-3Cx+Cx^2$
8

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}48=2B+3B+3C+C-4C+0+7B&\:\:\:\:\:\:\:(x=-1) \\ 26=C-4A+4B+2A-2B-5C&\:\:\:\:\:\:\:(x=1) \\ 59=-4A-5C+8A+10C+20B&\:\:\:\:\:\:\:(x=-2)\end{matrix}$
9

Proceed to solve the system of linear equations

$\begin{matrix}0A & + & 12B & + & 0C & =48 \\ -2A & + & 2B & - & 4C & =26 \\ 4A & + & 20B & + & 5C & =59\end{matrix}$
10

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & 12 & 0 & 48 \\ -2 & 2 & -4 & 26 \\ 4 & 20 & 5 & 59\end{matrix}\right)$
11

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & -5\end{matrix}\right)$
12

The decomposed integral equivalent is

$\int\left(\frac{1}{x-3}+\frac{4}{1+x}+\frac{-5}{x-2}\right)dx$
13

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int\frac{1}{x-3}dx+\int\frac{4}{1+x}dx+\int\frac{-5}{x-2}dx$
14

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|, where b=-3 and n=1 \ln\left|x-3\right|+\int\frac{4}{1+x}dx+\int\frac{-5}{x-2}dx 15 Apply the formula: \int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=1$ and $n=4$

$\ln\left|x-3\right|+4\ln\left|1+x\right|+\int\frac{-5}{x-2}dx$
16

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=-2$ and $n=-5$

$\ln\left|x-3\right|+4\ln\left|1+x\right|-5\ln\left|x-2\right|$
17

$\ln\left|x-3\right|+4\ln\left|1+x\right|-5\ln\left|x-2\right|+C_0$

$\ln\left|x-3\right|+4\ln\left|1+x\right|-5\ln\left|x-2\right|+C_0$

### Main topic:

Integrals by partial fraction expansion

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